sinxcosy=1/2,求cosx-siny范围。

取x=-y=π/4,
则sinx=cosy=√2/2,满足sinxcosy=1/2

此时，cosx=√2/2,siny=-√2/2,
cosx-siny=√2,

√2并不在你的答案：[－3/2,1/2]范围内。

所以，答案显然不正确

应该是：

设m=cosx-siny
m^2=(cosx)^2+(siny)^2-2cosxsiny
=2-(sinx)^2-(cosy)^2±2√[1-(sinx)^2][1-(cosy)^2]
=2-(sinx)^2-(cosy)^2±2√[1-(sinx)^2-(cosy)^2+(sinxcosy)^2]
=2-(sinx)^2-(cosy)^2±√[5-4(sinx)^2-4(cosy)^2]
设t=√[5-4(sinx)^2-4(cosy)^2]<=√(5-8sinxcosy)=1
∴0<=t<=1(当sinx=cosx,t=1)
f(t)=m^2=t^2/4±t+3/4<=t^2/4+t+3/4
=1/4(t+2)^2-1/4=g(t)
当t>-2,g(t)单调递增
f(t)<=g(t)<=g(1)=2
∴-√2<=m<=√2
∴cosx-siny∈[-√2,√2]

sinx*cosy=1/2
cosx*siny=t
两个式子相加,得到
sinxcosy+cosxsiny=1/2+t
sin(x+y)=1/2+t
|sin(x+y)|<=1
|1/2+t|<=1
-3/2<=t<=1/2
两个式子相减,得到
sinxcosy-cosxsiny=1/2-t
sin(x-y)=1/2-t
|sin(x-y)|<=1
|1/2-t|<=1
-1/2<=t<=3/2
综合得到-1/2<=t<=1